Question

Two particles $A$ and $B$ are projected with same speed so that the ratio of their maximum heights reached is $3:1$ . If the speed of $A$ is doubled without altering other parameters, the ratio of the horizontal ranges attained by $A$ and $B$ is

• A. $1:1$
• B. $2:1$
• C. $4:1$
• D. $3:2$

Answer 1

Answer: (C)

$4:1$

Answer 2

$H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ or $\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{1}}}{{{u}^{2}}{{\sin }^{2}}{{\theta }_{2}}}$
Or $\frac{3}{1}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}$
or $\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}=\frac{\sqrt{3}}{1}$
Logically, we can conclude that ${{\theta }_{1}}={{60}^{o}},{{\theta }_{2}}={{30}^{o}}$
Again $R=\frac{{{u}^{2}}\sin 2\theta }{g}$
Or $\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4{{u}^{2}}\sin 2{{\theta }_{1}}}{{{u}^{2}}\sin 2{{\theta }_{2}}}$
Or $\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4\sin 2({{60}^{o}})}{\sin 2(30{}^\circ )}=\frac{4\sin 120{}^\circ }{\sin 60{}^\circ }$
Or $\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4\times \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}=4$